Question

assuming that Huawei produces 5% of their total production of smartphones are defective. If 10 items are to be chosen at random from the production line, what then is the probability that;

a. all smartphones are not defective?

b. at most 2 of the smartphones are defective?

c. at least 3 of the smartphones are defective?

Answer 1

Let $X=$ the number of defective smartphones: $X\sim Bin(n, p).$

Given $n=10, p=0.05, q=0.95.$

a.

$$P(X=0)=\dbinom{10}{0}(0.05)^0(0.95)^{10-0}$$

$$=0.59873693924$$

b.

$$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$$

$$=\dbinom{10}{0}(0.05)^0(0.95)^{10-0}$$

$$+\dbinom{10}{1}(0.05)^1(0.95)^{10-1}+\dbinom{10}{2}(0.05)^2(0.95)^{10-2}$$

$$=0.59873693924+0.31512470486$$

$$+0.07463479852=0.98849644262$$

c.

$$P(X\ge 3)=1-P(X=0)-P(X=1)$$

$$-P(X=2)=1-\dbinom{10}{0}(0.05)^0(0.95)^{10-0}$$

$$-\dbinom{10}{1}(0.05)^1(0.95)^{10-1}-\dbinom{10}{2}(0.05)^2(0.95)^{10-2}$$

$$=1-0.59873693924-0.31512470486$$

$$-0.07463479852=0.01150355738$$